What is the unit vector that is normal to the plane containing #( i +k)# and #(i+2j+2k) #?
1 Answer
Explanation:
The vector we're looking for is
Using this fact, we can make a system of equations:
#vecn * (i+0j+k) = 0#
#(ai+bj+ck)(i+0j+k)=0#
#a+c = 0#
#vecn * (i+2j+2k) = 0#
#(ai+bj+ck) * (i+2j+2k) = 0#
#a+2b+2c = 0#
Now we have
#a+c = a+2b+2c#
#0 = 2b+c#
#therefore a+c = 2b+c#
#a = 2b#
#a/2 = b#
Now we know that
#ai + a/2j-ak#
Finally, we need to make this a unit vector, meaning we need to divide each coefficient of the vector by its magnitude. The magnitude is:
#|vecn|=sqrt(a^2+(a/2)^2+(-a)^2)#
#|vecn|=sqrt(9/4a^2)#
#|vecn| = 3/2a#
So our unit vector is:
#vecn = a/(3/2a)i + (a/2)/(3/2a)j + (-a)/(3/2a)k#
#vecn = 2/3i + 1/3j -2/3k#
Final Answer